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3 years ago

Simplify (3-6i) (-5+ i)

Mathematics

1 answer:

Tanya [424]3 years ago

5 0

Answer:

Simplifed using FOIL: -15+30t+(3-6t)i

Simplifed using distributive property: 15+30t+(3-6t)i

Step-by-step explanation:

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Josh is making 3 recipes one requires 2/3 cups of flour and otherwise 7/8 of a cup of flour in the third requires 3/4 cup of flo

kati45 [8]

Answer:

2 7/24C

Step-by-step explanation:

Start by finding the common denominator. In this case, it's 24. Next, start translation.

2/3C = 16/24C

7/8C = 21/24C

3/4C = 18/24C

Then find the sum.

16/24C + 21/24C + 18/24C = 55/24C = 2 7/24C

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3 years ago

A rainstorm produced a rainfall of 2 inches per hour how many hours would it take to get a rainfall amount of 1 foot

Elis [28]

There are 12 inches in one foot. If it rains 2 inches per hour and we need one foot we need to look for a number that can be multiplied by 2 to get 12.
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The answer is 6

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3 years ago

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In the typing World 120 words in 3 minutes is average how many words could they type in 2 hours

Molodets [167]

Answer:

7,200 words

Step-by-step explanation:

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2 years ago

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Use function notation to write the equation of the line

blondinia [14]

$f(x)=mx+b\to y=mx+b$

We have two points:

$(-3,\ 0);\ (-4,\ 7)$

The point slope form of the line:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute

$m=\dfrac{7-0}{-4-(-3)}=\dfrac{7}{-1}=-7$

$y-0=-7(x-(-3))\\\\y=-7(x+3)\\\\y=-7x-21$

Answer:

$f(x)=-7x-21$

8 0

2 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

2 years ago